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Boundaryless D4 Source Current Is Closed

Physical scope

Distributional closure of the oriented degree-five current for a smooth boundaryless D4 worldvolume, using compactly supported test forms; a worldvolume boundary adds its boundary current.

Statement

Let $X^{10}$ be oriented and $ι : W_{5} ↪ X$ a smooth proper oriented embedding. Define the degree-five current by $⟨ δ_{5} (W), η_{5} ⟩ : = \int_{W} ι^{*} η_{5}$ for every compactly supported smooth five-form $η_{5}$ ; in the boundaryless Nil case this is the current of D4 Distributional Source Current. For a degree- $k$ current $T$ , define its exterior derivative by $⟨ d T, β_{9 - k} ⟩ : = (- 1)^{k + 1} ⟨ T, d β_{9 - k} ⟩$ . If $\partial W = \emptyset$ , then $d δ_{5} (W) = 0$ . If $\partial W \neq \emptyset$ , assume ${A C}_{ω}$ , the countable-choice principle that every sequence of nonempty sets has a choice function, and orient $\partial W$ by the outward-normal-first convention; define $δ_{6} (\partial W)$ by $⟨ δ_{6} (\partial W), β_{4} ⟩ : = \int_{\partial W} (ι ∣_{\partial W})^{*} β_{4}$ for compactly supported smooth four-forms $β_{4}$ . Then $d δ_{5} (W) = δ_{6} (\partial W)$ . The boundary statement is conditional on this displayed countable-choice premise.

Facts & Assumptions

Given: The smooth proper oriented embedding, the current pairings and derivative convention in the Statement, and compactly supported test forms.

[F1]

D4 Distributional Source Current: in the boundaryless Nil case, $⟨ δ_{5} (W), η_{5} ⟩ = \int_{W} ι^{*} η_{5}$ ; the boundary-current extension used here is defined explicitly in the Statement.

[F2]

Finite chart localization gives choice-free integration and compact Stokes: on an oriented boundaryless manifold, the integral of a compactly supported exact top form is zero.

[F3]

The general Stokes theorem: under ${A C}_{ω}$ , for a smooth manifold with boundary and a compactly supported form of degree one below its dimension, $\int_{W} d η = \int_{\partial W} j^{*} η$ with outward-normal-first orientation.

[F4]

The exterior derivative commutes with pullback: for every smooth map $ι$ and smooth form $β$ , $d (ι^{*} β) = ι^{*} (d β)$ .

Proof

1.1givenF4algebra

Let $β_{4}$ be an arbitrary compactly supported smooth test four-form on $X$ . Since $ι$ is proper, $ι^{*} β_{4}$ is compactly supported on $W$ . By the current pairing and derivative convention in the Statement and [F4], $⟨ d δ_{5} (W), β_{4} ⟩ = (- 1)^{6} \int_{W} ι^{*} (d β_{4}) = \int_{W} d (ι^{*} β_{4})$ .

2.1givenF2step 1.1

If $\partial W = \emptyset$ , [F2] makes the last integral zero, so every test pairing of $d δ_{5} (W)$ vanishes. Thus $d δ_{5} (W) = 0$ distributionally in the boundaryless case.

3.1givenF3step 1.1∎

If $\partial W \neq \emptyset$ , [F3] gives $⟨ d δ_{5} (W), β_{4} ⟩ = \int_{\partial W} (ι ∣_{\partial W})^{*} β_{4} = ⟨ δ_{6} (\partial W), β_{4} ⟩$ . The equality for every test form is exactly $d δ_{5} (W) = δ_{6} (\partial W)$ , under [F3]'s stated ${A C}_{ω}$ premise.

Boundaryless D4 Source Current Is Closed

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources